Calculus of Real and Complex Variables

8.5 Spherical Coordinates In p Dimensions

Sometimes there is a need to deal with spherical coordinates in more than three dimensions. In this section,
this concept is defined and formulas are derived for these coordinate systems. Recall polar coordinates are
of the form

y1 = ρcosθ
y2 = ρsin θ

where ρ > 0 and θ ∈ ℝ. Thus these transformation equations are not one to one but they are one to one on

(0,∞ )

× [0,2π). Here I am writing ρ in place of r to emphasize a pattern which is about to
emerge. I will consider polar coordinates as spherical coordinates in two dimensions. I will also
simply refer to such coordinate systems as polar coordinates regardless of the dimension. This
is also the reason I am writing y1 and y2 instead of the more usual x and y. Now consider
what happens when you go to three dimensions. The situation is depicted in the following
picture.

PICT

From this picture, you see that y3 = ρcosϕ1. Also the distance between

(y1,y2)

and

(0,0)

is ρsin

(ϕ1)

.
Therefore, using polar coordinates to write

(y1,y2)

in terms of θ and this distance,

y1 = ρsinϕ1cosθ,
y2 = ρsinϕ1sin θ,
y3 = ρcosϕ1.

where ϕ1∈ ℝ and the transformations are one to one if ϕ1 is restricted to be in

[0,π]

. What was done is to
replace ρ with ρsinϕ1 and then to add in y3 = ρcosϕ1. Having done this, there is no reason to stop with
three dimensions. Consider the following picture:

PICT

From this picture, you see that y4 = ρcosϕ2. Also the distance between

Continuing this way, given spherical coordinates in ℝp, to get the spherical coordinates in ℝp+1, you let
yp+1 = ρcosϕp−1 and then replace every occurance of ρ with ρsinϕp−1 to obtain y1

⋅⋅⋅

yp in terms of
ϕ1,ϕ2,

⋅⋅⋅

,ϕp−1,θ, and ρ.

It is always the case that ρ measures the distance from the point in ℝp to the origin in ℝp, 0. Each
ϕi∈ ℝ and the transformations will be one to one if each ϕi∈

(0,π)

, and θ ∈

(0,2π)

. Denote by
hp

( ⃗ )
ρ,ϕ,θ

the above transformation.

It can be shown using math induction and geometric reasoning that these coordinates map
∏i=1p−2

(0,π)

×

(0,2π)

×

(0,∞)

one to one onto an open subset of ℝp which is everything except for the
set of measure zero Ψp

(N )

where N results from having some ϕi equal to 0 or π or for ρ = 0 or for θ equal
to either 2π or 0. Each of these are sets of Lebesgue measure zero and so their union is also a set of
measure zero. You can see that hp

(∏ )
pi−=21 (0,π)× (0,2π )× (0,∞ )

omits the union of the coordinate axes
except for maybe one of them. This is not important to the integral because it is just a set of measure
zero.

Theorem 8.5.1Let y = hp

( )
⃗ϕ,θ,ρ

be the spherical coordinate transformations in ℝp. Then lettingA = ∏i=1p−2

Proof: Formula 8.11 is obvious from the definition of the spherical coordinates because in the matrix of
the derivative, there will be a ρ in p− 1 columns. The first claim is also clear from the definition and math
induction or from the geometry of the above description. It remains to verify 8.12 and 8.13. It is clear hp
maps A× [0,∞) onto ℝp. Since hp is differentiable, it maps sets of measure zero to sets of measure zero.
Then

Now the claim about f ∈ L1 follows routinely from considering the positive and negative parts of the real
and imaginary parts of f in the usual way. ■

Note that the above equals

∫ ( ( )) ( )
f hp ⃗ϕ,θ,ρ ρp−1Φ ⃗ϕ,θ dmp
A¯× [0,∞ )

and the iterated integral is also equal to

∫ ∫ ( ( )) ( )
ρp−1 f hp ⃗ϕ,θ,ρ Φ ⃗ϕ,θ dmp −1dm
[0,∞ ) ¯A

because the difference is just a set of measure zero.

Notation 8.5.2Often this is written differently. Note that from the spherical coordinate formulas,f

( ( ))
h ⃗ϕ,θ,ρ

= f

(ρω )

where

|ω |

= 1. Letting Sp−1denote the unit sphere,

{ω ∈ ℝp : |ω | = 1}

, theinside integral in the above formula is sometimes written as

∫
f (ρω)dσ
Sp−1

where σ is a measure on Sp−1. See [80]for another description of this measure. It isn’t an important issuehere. Either 8.13or the formula

∫ ∞ ( ∫ )
ρp−1 f (ρω)dσ dρ
0 Sp−1

will be referred to as polar coordinates and is very useful in establishing estimates. Hereσ

( )
Sp−1

≡∫AΦ

( )
⃗ϕ,θ

dmp−1.

Example 8.5.3For what values of s is the integral∫B

(0,R)

( )
1 + |x|2

sdy bounded independent ofR? Here B

(0,R )

is the ball,

p
{x ∈ ℝ : |x| ≤ R}

.

I think you can see immediately that s must be negative but exactly how negative? It turns out it
depends on p and using polar coordinates, you can find just exactly what is needed. From the polar
coordinates formula above,